Problem: $f(x, y) = x(y - \sin(y))$ $\dfrac{\partial^2 f}{\partial y^2} = $
Taking a second order partial derivative is like taking a regular second order derivative. We take the partial derivative once, then we take another partial derivative. $\dfrac{\partial^2 f}{\partial y^2} = \dfrac{\partial}{\partial y} \left[ \dfrac{\partial f}{\partial y} \right]$ Let's differentiate! $\begin{aligned} \dfrac{\partial^2 f}{\partial y^2} &= \dfrac{\partial}{\partial y} \left[ \dfrac{\partial}{\partial y} \left[ x(y - \sin(y)) \right] \right] \\ \\ &= \dfrac{\partial}{\partial y} \left[ x - x\cos(y) \right] \\ \\ &= 0 + x \sin(y) \end{aligned}$ Therefore, $\dfrac{\partial^2 f}{\partial y^2} = x \sin(y)$.